# The Stacks Project

## Tag 0629

Lemma 15.26.8. Let $R$ be a ring. Let $f_1, \ldots, f_r$ be a sequence of elements of $R$. The complex $K_\bullet(f_1, \ldots, f_r)$ is isomorphic to the cone of the map of complexes $$f_r : K_\bullet(f_1, \ldots, f_{r - 1}) \longrightarrow K_\bullet(f_1, \ldots, f_{r - 1}).$$

Proof. Special case of Lemma 15.26.7. $\square$

The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5607–5618 (see updates for more information).

\begin{lemma}
\label{lemma-cone-koszul}
Let $R$ be a ring. Let $f_1, \ldots, f_r$ be a sequence of elements
of $R$. The complex $K_\bullet(f_1, \ldots, f_r)$ is isomorphic to the
cone of the map of complexes
$$f_r : K_\bullet(f_1, \ldots, f_{r - 1}) \longrightarrow K_\bullet(f_1, \ldots, f_{r - 1}).$$
\end{lemma}

\begin{proof}
Special case of
Lemma \ref{lemma-cone-koszul-abstract}.
\end{proof}

Comment #2447 by Raymond Cheng on March 9, 2017 a 2:42 am UTC

Should be $f_r \colon K_\bullet(f_1,\ldots,f_{r-1}) \to K_\bullet(f_1,\ldots,f_{r-1})$ instead of $f_n\colon K_\bullet(f_1,\ldots,f_{r-1}) \to K_\bullet(f_1,\ldots,f_{r-1})$.

Comment #2489 by Johan (site) on April 13, 2017 a 10:57 pm UTC

Thanks, fixed here.

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